APPM 7400 Lesson 11: Spatial Poisson Processes...The Spatial Poisson Process Consider a spatial con...

Stochastic SimulationAPPM 7400

Lesson 11: Spatial Poisson Processes

October 3, 2018

Lesson 11: Spatial Poisson Processes Stochastic Simulation October 3, 2018 1 / 24

The Spatial Poisson Process

Consider a spatial configuration of points in the plane:

Notation:

Let S be a subset of R2. (Rk)(Assume S is normalized to have volume 1.)

Let A be the family of all subsets of S .

For A ∈ A, let |A| denote the size of A. (length, area, volume,...)

Let N(A) be the number of points in the set A.

Notation:

Then {N(A)}A∈A is a homogenous spatial Poisson process with intensityλ > 0 if:

For each A ∈ A, N(A) ∼ Poisson(λ|A|).

For every finite collection A1,A2, . . . ,An of disjoint subsets of S ,

N(A1),N(A2), . . . ,N(An)

are independent.

(N(∅) = 0)

Alternatively, a spatial Poisson process satisfies the following axioms:

i. If A1,A2, . . . ,An are disjoint regions in S , then

N(A1),N(A2), . . . ,N(An)

are independent random variables and

N(A1 ∪ A2 ∪ · · · ∪ An) = N(A1) + N(A2) + · · ·+ N(An)

ii. The probability distribution of N(A) depends on the set A onlythrough its size |A|.

N(A1),N(A2), . . . ,N(An)

N(A1 ∪ A2 ∪ · · · ∪ An) = N(A1) + N(A2) + · · ·+ N(An)

N(A1),N(A2), . . . ,N(An)

N(A1 ∪ A2 ∪ · · · ∪ An) = N(A1) + N(A2) + · · ·+ N(An)

iii. There exists a λ such that

P(N(A) ≥ 1) = λ|A|+ o(|A|)

iv. There is probability zero of points overlapping:

P(N(A) ≥ 2) = o(|A|)

iii. There exists a λ such that

P(N(A) ≥ 1) = λ|A|+ o(|A|)

iv. There is probability zero of points overlapping:

P(N(A) ≥ 2) = o(|A|)

If these axioms are satisfied, we have:

P(N(A) = k) =e−λ|A|(λ|A|)k

for k = 0, 1, 2, . . .

Consider a subset A of S :

There are 3 points in A... how are they distributed within A?Expect a uniform distribution...

There are 3 points in A... how are they distributed within A?

Expect a uniform distribution...

There are 3 points in A... how are they distributed within A?Expect a uniform distribution...Lesson 11: Spatial Poisson Processes Stochastic Simulation October 3, 2018 8 / 24

In fact, for any B ⊆ A, we have

P(N(B) = 1|N(A) = 1) =|B||A|

Proof:

P(N(B) = 1|N(A) = 1) =P(N(B) = 1,N(A) = 1)

P(N(A) = 1)

= P(N(B)=1,N(A∩B′)=0)P(N(A)=1)

= λ|B|e−λ|B|·e−λ|A∩B′|

λ|A|eλ|A| = |B||A|

P(N(B) = 1|N(A) = 1) =|B||A|

Proof:

P(N(B) = 1|N(A) = 1) =P(N(B) = 1,N(A) = 1)

P(N(A) = 1)

= P(N(B)=1,N(A∩B′)=0)P(N(A)=1)

λ|A|eλ|A| = |B||A|

P(N(B) = 1|N(A) = 1) =|B||A|

Proof:

P(N(B) = 1|N(A) = 1) =P(N(B) = 1,N(A) = 1)

P(N(A) = 1)

= P(N(B)=1,N(A∩B′)=0)P(N(A)=1)

λ|A|eλ|A| = |B||A|

P(N(B) = 1|N(A) = 1) =|B||A|

Proof:

P(N(B) = 1|N(A) = 1) =P(N(B) = 1,N(A) = 1)

P(N(A) = 1)

= P(N(B)=1,N(A∩B′)=0)P(N(A)=1)

λ|A|eλ|A|

= |B||A|

P(N(B) = 1|N(A) = 1) =|B||A|

Proof:

P(N(B) = 1|N(A) = 1) =P(N(B) = 1,N(A) = 1)

P(N(A) = 1)

= P(N(B)=1,N(A∩B′)=0)P(N(A)=1)

λ|A|eλ|A| = |B||A|

Simulating a spatial Poisson pattern over a rectangular regionS = [a, b]× [c , d ]:

simulate a Poisson number of points

scatter that number of points uniformly over S

ie: For each point, draw U1,U2 indep. unif (0, 1)’s and place it at

((b − a)U1 + a, ((d − c)U2 + c))

Generalization of the uniform result:

For any B ⊆ A, we have

N(B)|N(A) = n ∼ bin(n, |B|/|A|)

Generalization of the uniform result:

For any B ⊆ A, we have

N(B)|N(A) = n ∼ bin(n, |B|/|A|)

More Generalization:

For disjoint subsets A1,A2, . . . ,Am ⊆ A,

P(N(A1) = n1,N(A2) = n2, . . . ,N(Am) = nm|N(A) = n)

n1!n2! · · · nm!

(|A1||A|

·(|A2||A|

· · ·(|Am||A|

for n1 + n2 + · · ·+ nm = n.

More Generalization:

For disjoint subsets A1,A2, . . . ,Am ⊆ A,

P(N(A1) = n1,N(A2) = n2, . . . ,N(Am) = nm|N(A) = n)

n1!n2! · · · nm!

(|A1||A|

·(|A2||A|

· · ·(|Am||A|

for n1 + n2 + · · ·+ nm = n.

Consider a two-dimensional spatial Poisson process of particles in the planewith intensity parameter λ.

Let’s determine the (random) distance D between a particle and it’snearest neighbor.

For x > 0,

FD(x) = P(D ≤ x) = 1− P(D > x)

= 1− P( no other particles in disk with areaπx2 centered at the particle )

= 1− e−λπx2

For x > 0,

FD(x) = P(D ≤ x) = 1− P(D > x)

= 1− e−λπx2

For x > 0,

FD(x) = P(D ≤ x) = 1− P(D > x)

= 1− e−λπx2

For x > 0,

FD(x) = P(D ≤ x) = 1− P(D > x)

= 1− e−λπx2

For x > 0,

FD(x) = P(D ≤ x) = 1− P(D > x)

= 1− e−λπx2

fD(x) =d

dxFD(x) = 2λπxe−λπx

for x > 0. (Weibull distribution!)

Similarly, in 3-D:

FD(x) = 1− e−λ4π3x3

fD(x) =d

dxFD(x) = 4πλx2e−λ

4π3x3

(Weibull distribution!)

fD(x) =d

dxFD(x) = 2λπxe−λπx

for x > 0. (Weibull distribution!)

Similarly, in 3-D:

FD(x) = 1− e−λ4π3x3

fD(x) =d

dxFD(x) = 4πλx2e−λ

4π3x3

(Weibull distribution!)

Example: Spatial Patterns in Statistical Ecology

Consider a wide expanse of open ground of a uniform character.(example: muddy bed of a recently drained lake)

The number of wind-dispersed seeds occurring in any particular“quadrat” on this surface is well modeled by a Poisson randomvariable.

The reason this tends to be true is due to the Poisson approximationto the binomial distribution which will hold if there are many seedswith an extremely small chance of falling into the quadrat.

Suppose now that the probability that a seed germinates is p and thatthey are not sufficiently packed together to interact at this stage.

Question: What is the distribution of the number of germinated seeds?

Answer: This is a thinned spatial Poisson process with intensity pλ.

(So, the surviving seedlings continue to be distributed “at random”.)

Simulation Problem:

Two types of seeds are randomly dispersed on a one-acre fieldaccording to two independent spatial Poisson processes withintensities λ1 and λ2.

Type 1 and Type 2 seeds will germinate with probabilities p1 and p2,respectively.

Type 1 plants will produce K offshoot plants on runners randomlyspaced around the plant where K ∼ geom(p). (P(K = 0) = p)

Suppose that time is discretized as follows:

Time 0: seeds are dispersedTime 1: seeds germinateTime 2: offshoot plants produced

Suppose that the one-acre field is evenly divided into 10× 10quadrats.

Simulation Problem:

Assume that the number of offshoot plants that fall into a quadratdifferent from their parent plans is negligible.

A particular insect population can only be supported if at least 75%of the quadrats contain at least 35 plants.

Using p = 0.9, p1 = 0.7, and p2 = 0.8, explore the values of λ1 andλ2 that will give the insect population a 95% chance of surviving.

Use the hugely simplifying assumption that there is no “real time”component of this process. (In particular, assume that offshoot plantsdo not have further offshoots.)

Simulation Problem:

Tips on Simulating This:

Keep in mind that we don’t really have to keep track of whereindividual plants are, only the number in each quadrat.

Note that we don’t have to consider germination of the plants as asecond step after the arrival of the seeds– instead onsider a thinnedspatial Poisson number of plants of type i with intensity piλi .

Rather than drawing uniformly distributed locations for the seeds, wecan simulate the number for each quadrat separately (and ignorelocations) using the fact that each quadrat contains aPoisson(piλi/100) number of germinating seeds.

How to deal with offshoot plants...

It would be nice if we could further modify the Poisson number ofseeds for Type 1 plants.

We can’t. :(

We can, however, simplify the generation of offshoot plants, dealingwith all plants in a particular quadrat together by adding a negativebinomial number of plants to each quadrat.

We can’t. :(

Specifics of my simulation:

created two 10× 10 arrays A1 and A2

filled the arrays by drawing 100 values from the Poisson(p1λ1/100)distribution and 100 values from the Poisson(p2λ2/100) distribution

went through the first array and replaced A1[i , j ] with A1[i , j ] + Ki ,j

where the Ki ,j are drawn independently from the negative binomialdistribution with r = A1[i , j ] and p = 0.9

let A = A1 + A2 and determined the proportion of entries in A with35 or more plants

Repeat... Want to see a proportion of 0.75 or greater inapproximately 95% of simulations

Explore by changing the intensities (the λ’s)

Depending on the efficiency of your simulation, this could be timeconsuming, so you might think about choosing λ’s in the right ball park.

Ignoring offshoot plants, we know to expect

p1λ1/100 + p2λ2/100

in each quadrat.

p1λ1/100 + p2λ2/100

in each quadrat.

p1λ1/100 + p2λ2/100

in each quadrat.

p1λ1/100 + p2λ2/100

in each quadrat.

I started with λ1 = λ2 = 2500. (Then p1λ1/100 + p2λ2/100 = 37.5.)

Results for some simulations:

Sim Prop. of Quadrats Containing Support?at Least 35 Plants

1 0.72 No2 0.75 Yes3 0.82 Yes4 0.83 Yes5 0.81 Yes6 0.79 Yes7 0.71 No

(Continuing on for 10, 000 simuations, I found that the popoulation couldbe supported using these λ’s roughly 87% of the time.)

I started with λ1 = λ2 = 2500. (Then p1λ1/100 + p2λ2/100 = 37.5.)Results for some simulations:

APPM 7400 Lesson 11: Spatial Poisson Processes...The Spatial Poisson Process Consider a spatial con...

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